Determine the depth of embedment for the sheet-piling shown in Fig. Ex. 20.la by rigorous analysis. Determine also the minimum section modulus. Assume an allowable flexural stress f_{b}=175 MN / m ^{2}. The soil has an effective unit weight of 17 kN / m ^{3} and angle of internal friction of 30°.
Question 20.1: Determine the depth of embedment for the sheet-piling shown ...
For \phi=30^{\circ}, K_{A}=\tan ^{2}\left(45^{\circ}-\phi / 2\right)=\tan ^{2} 30=\frac{1}{3}
K_{p}=\frac{1}{K_{A}}=3, K=K_{P}-K_{A}=3-\frac{1}{3}=2.67
The pressure distribution along the sheet pile is assumed as shown in Fig. Ex. 20.1(b)
\bar{p}_{a}=\gamma H K_{A}=17 \times 6 \times \frac{1}{3}=34 kN / m ^{2}
From Eq (20.1)
y_{0}=\frac{\gamma H K_{A}}{\gamma\left(K_{p}-K_{A}\right)}=\frac{\bar{p}_{a}}{\gamma K} (20.1)
\begin{aligned}&y_{0}=\frac{\bar{p}_{a}}{\gamma\left(K_{p}-K_{A}\right)}=\frac{34}{17 \times 2.67}=0.75 m. \\&P_{a}=\frac{1}{2} \bar{p}_{a} H+\frac{1}{2} \bar{p}_{a} y_{0}=\frac{1}{2} \times 34 \times 6+\frac{1}{2} \times 34 \times 0.75\end{aligned}
= 102 + 12.75 = 1 14.75 kN/meter length of wall or say 1 15 kN/m.
\begin{aligned}&\bar{p}_{p}=\gamma D\left(K_{p}-K_{A}\right)-\bar{p}_{a}=17 \times D \times 2.67-34=45.4 D-34 \\&\bar{p}_{p}^{\prime}=\gamma H K_{p}+\gamma D\left(K_{p}-K_{A}\right)=17 \times 6 \times 3+17 \times D \times 2.67=306+45.4 D \\&\bar{p}_{p}^{\prime \prime}=\gamma H K_{p}+\gamma_{0}\left(K_{p}-K_{A}\right)=17 \times 6 \times 3+17 \times 0.75 \times 2.67=340 kN / m ^{2}\end{aligned}
To find \bar{y}
\begin{aligned}&P_{a} \bar{y}=\frac{1}{2} \bar{p}_{a} H \quad \frac{H}{3}+y_{0} \quad+\frac{1}{2} \bar{p}_{a} y_{0} \quad \frac{2}{3} y_{0} \\&=\frac{1}{2} \times 34 \times 6 \times(2+0.75)+\frac{1}{2} \times 34 \times 0.75 \times \frac{2}{3} \times 0.75=286.9\end{aligned}
Now D_{0} can be found from Eq. (20.4), namely
D_{0}^{4}+C_{1} D_{0}^{3}+C_{2} D_{0}^{2}+C_{3} D_{0}+C_{4}=0 (20.4)
D_{0}^{4}+C_{1} D_{0}^{3}+C_{2} D_{0}^{2}+C_{3} D_{0}+C_{4}=0
\begin{aligned}&C_{1}=\frac{\bar{p}_{p}^{\prime \prime}}{\gamma K}=\frac{340}{17 \times 2.67}=7.49, \quad C_{2}=-\frac{8 P_{a}}{\gamma K}=-\frac{8 \times 115}{17 \times 2.67}=-20.3 \\&C_{3}=-\frac{6 P_{a}}{(\gamma K)^{2}}\left(2 \bar{y} \gamma K+\bar{p}_{p}^{\prime \prime}\right) \quad=-\frac{6 \times 115}{(17 \times 2.67)^{2}}(2 \times 2.50 \times 17 \times 2.67+340)=-189.9 \\&C_{4}=-\frac{6 P_{a} \bar{y} \bar{p}_{p}^{\prime \prime}+4 P_{a}^{2}}{(\gamma K)^{2}}=-\frac{6 \times 115 \times 2.50 \times 340+4 \times(115)^{2}}{(17 \times 2.67)^{2}}=-310.4\end{aligned}
Substituting for C_{1}, C_{2}, C_{3} \text { and } C_{4}, and simplifying we have
D_{0}^{4}+7.49 D_{0}^{3}-20.3 D_{0}^{2}-189.9 D_{0}-310.4=0
This equation when solved by the method of trial and error gives
D_{0} \approx 5.3 m
Depth of Embedment
D=D_{0}+y_{0}=5.3+0.75=6.05 m
Increasing D by 40%, we have’
D(\text { design })=1.4 \times 6.05=8.47 m \text { or say } 8.5 m .
(c) Section modulus
From Eq. (20.6) (The point of zero shear)
\bar{y}_{0}=\sqrt{\frac{2 P_{a}}{\gamma K}} (20.6)
\begin{aligned}&\bar{y}_{o}=\sqrt{\frac{2 P_{a}}{\gamma K}}=\sqrt{\frac{2 \times 115}{17 \times 2.67}}=2.25 m \\&M_{\max }=P_{a}\left(\bar{y}+\bar{y}_{o}\right)-\frac{1}{6} \bar{y}_{o}^{3} \gamma K \\&=115(2.50+2.25)-\frac{1}{6}(2.25)^{3} \times 17 \times 2.67 \\&=546.3-86.2=460 kN – m / m\end{aligned}
From Eq. (20.8)
Z _{s}=\frac{M_{\max }}{f_{b}} (20.8)
Section modulus
Z_{s}=\frac{M_{\max }}{f_{b}}=\frac{460}{175 \times 10^{3}}=26.25 \times 10^{-2} m ^{3} / m \text { of wall }
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